logseq_notes/pages/OJ notes/pages/Leetcode Binary-Search.md

Leetcode Binary-Search

2022-07-09 09:34

Algorithms:

#algorithm #binary_search

Data structures:

#DS #array

Difficulty:

#coding_problems #difficulty_easy

Additional tags:

#leetcode #CS_list_need_practicing

Revisions:

N/A

Related topics:

Links:

• Link to problem

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Problem

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Examples

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1

Constraints

• 1 <= nums.length <= 104
• -104 < nums[i], target < 104
• All the integers in nums are unique.
• nums is sorted in ascending order.

Thoughts

[!summary] This is a #binary_search problem.

Key takeout:

int r = nums.size() - 1;

make sure r is never OOB (l == r && r = array.size())

Solution

==#TODO: write in recursion==

iteration

class Solution {
public:
int search(vector<int> &nums, int target) {
  // Why - 1?
  // Make sure mid is never OOB (l == r && r = size)
  int r = nums.size() - 1;
  int l = 0;
  int mid;
  int val;

  do {
    // l + ( r - l ) / 2 or (l + r) / 2 Both ok
    mid = (l + r) / 2;
    val = nums[mid];
    if (val == target) {
      return mid;
    } else if (val < target) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  } while (l <= r);

  return -1;
}
};